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Mathematica

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  • $\begingroup$ If you have $n$ arbitrary permutations, you must step through each before you have the final state. You can break this up in a number of ways (tracing the sequence of positions of each individual element, for instance), but you must still step through all in the arbitrary case. If, however, there are cycles or identity permutations or other non-random permutations, you can speed up the calculation. Then of course there are ways to parallelize the computation. $\endgroup$ Commented Mar 16, 2015 at 23:07
  • $\begingroup$ This may be useful: (6884708) $\endgroup$ Commented Mar 17, 2015 at 0:49
  • $\begingroup$ @Mr.Wizard I think the StackExchange question you linked concerns the ranking of permutations on $n$ distinct symbols. Ranking permutations on $n$ symbols, possibly with duplicates, is a rather different problem. $\endgroup$ Commented Mar 17, 2015 at 1:10
  • $\begingroup$ Okay, how about: (14212030) $\endgroup$ Commented Mar 17, 2015 at 1:14
  • $\begingroup$ @Mr.Wizard If I understand the accepted answer correctly, that (un)ranking function is precisely the one I've implemented in the nthPermutation code above. The problem is, that ranking function gives the permutations in lexicographic order, while the ranking function Mathematica uses internally is decidedly not lexicographic--examine the output of Permutations[{1, 3, 1, 2, 3}] and you'll see what I mean. $\endgroup$ Commented Mar 17, 2015 at 1:19